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Q. No.The root mean square speed of smoke particles of mass 5 × 10–17 kg in their Brownian motion in air at NTP is, approximately:
[Given k = 1.38 × 10–23 JK–1 ]
1. 60 mm s–1
2. 12 mm s–1
3. 15 mm s–1
4. 36 mm s–1
Subtopic: Types of Velocities |
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Level 2: 60%+
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\(0.056~\text{kg}\) of Nitrogen is enclosed in a vessel at a temperature of \(127^\circ \text C.\) The amount of heat required to double the speed of its molecules is:
(take \(R = 2 ~\text{cal mole}^{-1}\text{K}^{-1}\))
1. \(10~\text{Kcal}\)
2. \(12~\text{Kcal}\)
3. \(14~\text{Kcal}\)
4. \(16~\text{Kcal}\)
(take \(R = 2 ~\text{cal mole}^{-1}\text{K}^{-1}\))
1. \(10~\text{Kcal}\)
2. \(12~\text{Kcal}\)
3. \(14~\text{Kcal}\)
4. \(16~\text{Kcal}\)
Subtopic: Kinetic Energy of an Ideal Gas |
66%
Level 2: 60%+
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The relation between root mean square speed (vrms) and most probable speed (vp) for the molar mass M of oxygen gas molecule at the temperature of 300 K will be:
1. \(\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{2}{3}} \mathrm{v}_{\mathrm{p}} \)
2. \(\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3}{2}} \mathrm{v}_{\mathrm{p}} \)
3. \(\text v_{rms} = \text v_p\)
4. \(\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{1}{3}} \mathrm{v}_{\mathrm{P}}\)
1. \(\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{2}{3}} \mathrm{v}_{\mathrm{p}} \)
2. \(\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{3}{2}} \mathrm{v}_{\mathrm{p}} \)
3. \(\text v_{rms} = \text v_p\)
4. \(\mathrm{v}_{\mathrm{rms}}=\sqrt{\frac{1}{3}} \mathrm{v}_{\mathrm{P}}\)
Subtopic: Types of Velocities |
87%
Level 1: 80%+
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The ratio of specific heats \(\left (\dfrac{C_p}{C_v}\right)\) in terms of degree of freedom \((f)\) is given by:
| 1. | \(\left(1+\dfrac{f}{3}\right) \) | 2. | \(\left(1+\dfrac{2}{f}\right)\) |
| 3. | \(\left(1+\dfrac{f}{2}\right) \) | 4. | \(\left(1+\dfrac{1}{f}\right)\) |
Subtopic: Law of Equipartition of Energy |
88%
Level 1: 80%+
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When a gas filled in a closed vessel is heated by raising the temperature by \(1^\circ \text{C},\) its pressure increase by \(0.4\)%. The initial temperature of the gas is:
1. \(350\) K
2. \(150\) K
3. \(100\) K
4. \(250\) K
1. \(350\) K
2. \(150\) K
3. \(100\) K
4. \(250\) K
Subtopic: Ideal Gas Equation |
72%
Level 2: 60%+
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A flask contains argon and oxygen in the ratio of \(3:2\) in mass and the mixture is kept at \(27^\circ\mathrm{C}\). The ratio of their average kinetic energy per molecule respectively will be:
1. \(3:2\)
2. \(9:4\)
3. \(2:3\)
4. \(1:1\)
1. \(3:2\)
2. \(9:4\)
3. \(2:3\)
4. \(1:1\)
Subtopic: Kinetic Energy of an Ideal Gas |
74%
Level 2: 60%+
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A mixture of hydrogen and oxygen has a volume of \(2000\) cm3, temperature \(300\) K, pressure \(100\) kPa and mass
\(0.76\) g. The ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture will be:
\(0.76\) g. The ratio of the number of moles of hydrogen to the number of moles of oxygen in the mixture will be:
| 1. | \( 1 \over 3\) | 2. | \(3 \over 1\) |
| 3. | \( 1 \over 16\) | 4. | \( 16 \over 1\) |
Subtopic: Ideal Gas Equation |
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Level 3: 35%-60%
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For a perfect gas, two pressures \(P_1\) and \(P_2\) are shown in the figure. The graph shows:
1. \(P_1 > P_2\)
2. \(P_1 < P_2\)
3. \(P_1 = P_2\)
4. Insufficient data to draw any conclusion
1. \(P_1 > P_2\)
2. \(P_1 < P_2\)
3. \(P_1 = P_2\)
4. Insufficient data to draw any conclusion
Subtopic: Ideal Gas Equation |
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Level 1: 80%+
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According to the kinetic theory of gases,
Choose the most appropriate answer from the options given below:
| (A) | the motion of the gas molecules freezes at \(0^\circ\text C.\) |
| (B) | the mean free path of gas molecules decreases if the density of molecules is increased. |
| (C) | the mean free path of gas molecules increases if the temperature is increased keeping the pressure constant. |
| (D) | average kinetic energy per molecule per degree of freedom is \(\dfrac32k_B T\) (for monoatomic gases). |
| 1. | (A) and (C) only |
| 2. | (B) and (C) only |
| 3. | (A) and (B) only |
| 4. | (C) and (D) only |
Subtopic: Mean Free Path |
60%
Level 2: 60%+
Hints
The total internal energy of two moles of a monoatomic ideal gas at temperature \(T = 300~\text{K}\) will be:
(Given: \(R = 8.31\) J/mol.K)
1. \(4789~\text{J}\)
2. \(7479~\text{J}\)
3. \(5896~\text{J}\)
4. \(8346~\text{J}\)
(Given: \(R = 8.31\) J/mol.K)
1. \(4789~\text{J}\)
2. \(7479~\text{J}\)
3. \(5896~\text{J}\)
4. \(8346~\text{J}\)
Subtopic: Specific Heat |
77%
Level 2: 60%+
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